10. Applications of 1-D Hermite elements

Transcription

1 10. Applications of 1-D Hermite elements Introduction General case fourth-order beam equation Integral form Element Arrays C1 Element models Classic beams Structural symmetry The Rayleigh quotient Multiple span beams BOEF without axial load BOEF with axial load BOEF with axial load and end transverse force Summary Exercises Applications of 1-D Hermite elements 10.1 Introduction: Most one-dimensional applications of Hermite polynomials involve ODEs of even degree four and higher because their equivalent integral forms require the interelement continuity of at least the first derivative of the solution. A vast number of fourth-order ODEs for beam and frame (the bending extension of trusses) studies have been published. Since those structural elements are very common, many solutions of the fourth-order ODEs, with several different boundary conditions and/or various source (loading) terms are available in engineering handbooks. To introduce the engineering terminology, a beam is a one-dimensional structure intended to support loads (forces and moments) perpendicular to its axis. In other works, axial effects are not considered in a beam. A beam-column is a beam that is also subjected to axial displacements and forces (like those in Chapter 8). A frame is a structural system formed by connecting beam-columns in a non-collinear manner. Hermite interpolation can also be applied to lower order ODEs if and only if the coefficients and sources (material properties and loads) are constant or smoothly varying over the entire domain. Those applications can improve the accuracy of the result when using a small number of degrees of freedom. Also, some nonessential boundary conditions (NBC) can be exactly satisfied in the system matrices resulting from a Hermite interpolation model. That contrasts with C 0 interpolations where the NBC are only satisfied in a weak sense. All of the prior examples in Chapter 8 were based on the Galerkin method of weighted residuals. For completeness, the same ODE in Ex solved by the Galerkin method will be solved now by the Least Squares method of weighted residuals which is quite different. The least squares approach always yields a symmetric set of matrix equations (which is desirable), but it 1

2 also requires a higher level if interpolation continuity (which is difficult in higher dimensional physical spaces). Example * Given: Using the classic least squares method of weighted residuals, formulate the approximation of the ODE: d 2 u dx 2 + cx = 0, x =]0, L[ Then set c = L = 1 and apply the boundary conditions that u*(0)=0 and du*(l)/dx=0 and compare the result to the cubic exact solution u (x) = cl 2 x 2 c x 3 6. Solution: To employ classic least squares it is necessary to first define the residual error in terms of the unknown solution, u e. Approximate u (x) u(x) = H(x)u e and assume a linear geometry mapping of x(r) = x 1 + rl e so the Jacobian is constant. Then the residual error is R e (x) = d2 H(x) dx 2 u e + cx 0 The least squares weighting is defined as w k (x) = Re (x) e = d2 H k (x) u k dx 2, and w(x) T = d2 H T (x) dx 2 Then the weighted residual null vector is 0 L L I = w(x) T R e (x)dx = 0 = d2 H T (x) dx 2 ( d2 H(x) dx 2 u e + cx) dx 0 The highest derivative in this integral is two. To eventually replace the integral as one of an assembly of elements, calculus requires that the solution across element interfaces must be continuous to one derivative less. In other words, the interpolations for thus approach must have C 1 continuity between elements. That means the function and its slope must be continuous. Such interpolations are members of the Hermite polynomial family and the have u(x) and du(x) dx θ(x) as the nodal degrees of freedom (n g = 2). The element matrix system becomes L e [ d2 H T (x) d 2 L H (x) e dx 2 dx 2 L e dr] u e + { d2 H T (x) dx 2 cx L e dr} = {0} 0 The simplest element in that family is the two-node Hermite line element with two DOF per node. Thus, the element includes n n n g = n i = 2 2 = 4 independent DOFs. They define a cubic polynomial in one-dimensional space. The resulting interpolations were given previously in the Summary of Chapter 2 and are listed in the library function Hermite_1D_C1_library.m and are: u(x) = [(1 3r 2 + 2r 3 ) (r 2r 2 + r 3 )L e (3r 2 2r 3 ) (r 3 r 2 )L e ] { Therefore, the second derivative with respect to x is d 2 u dx 2 = d2 u dr 2 dr 2 dr 2 = d2 u 1 dr 2 (L e ) 2 = 1 d 2 H (r) (L e ) 2 dr 2 0 u 1 θ 1 u } 2 θ 2 e = H(x)u e 2

3 d 2 u dx 2 = 1 (L e ) 2 [( r) ( 4 + 6r)Le (6 12r) (6r 2)L e θ ] { 1 u } 2 θ 2 Therefore, substituting d 2 H (x) dx 2 into the above two integrals yields the governing matrix system, before enforcing the essential conditions as: 12 6L e 12 6L e u L e 3 [ 6L e 4L e 2 6L e 12 θ 12 6L e 12 6L e ] { 1 x u } = { 1 } 2 1 6L e 2L e 2 6L e 4L e 2 θ 2 L e (L e x 1 + 1) Note here that the quantities (du(0) dx θ 1 and du(l) dx θ 2 ) that are usually defined as the nonessential boundary conditions appear as nodal DOFs in the Hermite form and can be specifically enforced. That is because Hermite interpolation requires that the source (here cx) not be discontinuous, yet the Galerkin assumption does not have that restriction. Here the essential boundary condition u 1 = 0 = u(0) and the nonessential boundary condition of du(l) dx θ 2 = 0 can also be specified. Multiply columns 1 and 4 by their known values (an EBC and a NBC) and carry those known values to the right hand size. Then, only two independent equations remain in the second and third rows (the left slope and right solution value): 1 L e 3 [ 4Le2 6L e 6L e 4L e 2 ] {θ 1 } = { x 1 } u 1 u 2 1 L e 3 { 6Le 12 } θ 2 L e 3 { 12 6L e} = {x 1 } 1 Inverting gives the solution: { θ 1 } = Le u 2 6 { 3(Le + 2x 1 ) L e (2L e + 3x 1 ) } These nodal solution values are again analytically exact. For the numerical values of x 1 = 0 and L e = 1. { θ 1 u 2 } = 1 6 {3 2 } In this case the solution is exact everywhere because the cubic exact solution is included in the cubic interpolation. Example Given: Repeat Ex using the Galerkin method and the same Hermite element. Solution: From Ex. 8.x-x, the element matrices are 10.2 General case fourth-order beam equation: Figure shows the transverse displacement, v(x), of a beam on an elastic foundation. The beam is connected to a continuous series of foundation springs, of stiffness k per unit length. The other end of the foundation spring has a known displacement, v f, which is almost always zero. The differential equation of transverse equilibrium of a beam, of flexural stiffness EI, resting on a foundation, with a transverse load per unit length of w(x), subjected to a tensile axial load N is: u 1 e or d 2 [EI(x) d2 v dx 2 dx 2] d dv [N(x) ] + k(x)[v v dx dx ] f(x) = 0 3

4 d 2 [EI(x) d2 v dx 2 dx 2] N(x) d2 v dn(x) dv + k(x)[v v dx 2 dx dx ] f(x) = 0. (10.2-1) where v(x) is the transverse displacement of the beam. When the foundation effect is present the structure is usually called a beam on an elastic foundation, which is abbreviated as BOEF. The beam equation is a fourth-order ordinary differential equation. Therefore, it will generally need four boundary conditions. Figure A beam-column on an elastic foundation Each beam has a cross-section with two principle axes associated with the second-moments of inertia, I zz and I yy. For the coordinate system shown above the principle inertia is the former: I I zz = y 2 da. Its defining axis is the z-axis coming out of the paper and lies in the A horizontal plane. Conversely, the minor inertia y-axis lies in the vertical plane of the paper. Any changes in the two inertia axis directions becomes important if the beam is part of a space frame member located in general three-dimensional space, and they are discussed later. The distributed load per unit length can include point transverse shear loads, V, by using the Dirac Delta distribution. Likewise, employing a doublet distribution in defining f(x) allows for the inclusion of point couples, or moments, M. Related physical quantities are the slope (angle of rotation 1), θ(x) = v (x), the bending moment, M(x) = EIv (x), and the transverse shear force, V(x) = EIv (x). Engineers designing beams are usually interested in the last two quantities since they define the stress levels and the material failure criteria. That means that a model used in designing a beam should have accurate third derivatives, or a fine mesh. The sign conventions are that the position, x, is positive to the right, the deflection, v, point forces, P, and the line load, w, are positive in the y-direction (upward), and the slopes and moments are positive in the counter-clockwise direction. The axial force, N(x), is positive when it is in tension, which stabilizes the system, and negative when in compression, which de-stabilizes the system. Those two cases are typically called a tensioned-beam and a beam-column, respectively. When the axial force is in tension the solution to the homogeneous equation tends to smoothly decay away from the supports. Conversely, when the axial force is in compression the solution to the homogeneous solution oscillates and decays very slowly. Frequently, the axial force of interest is an unknown global constant to be computed. Then the model ODE becomes an eigen-problem to determine the axial force that causes buckling. Eigen-problems, including the buckling of beams are covered later in Chapter 12. The rate of change in the axial force, dn(x) dx, is usually small, especially over the length of a single element. Therefore, it could be neglected in some cases. However, if the axial force is a given function of x, then the gradient of N(x) needs to be retained in the approximate solution. As seen below, the retention of the dn(x) dx contribution leads to a non-symmetric set of system matrix equations. 4

5 In common drilling applications, or in structural piles, the x-axis may be along or near vertical. Then, the member weight per unit length, say w, must be included. Let β be the angle between the x-axis and the downward vertical direction. Then the rate of change of the axial force becomes dn dx = w cos β (10.2-2) Since the foundation springs are restrained by the displacement v at their end not connected to the beam, there can be no rigid body motion of the beam. In other words, the assembled equations should never be singular and it is acceptable to just have nonessential boundary conditions applied to the beam. A simple Winkler foundation model like this one can push or pull on the beam as needed and no gaps can occur. If a foundation is not present, then enough essential boundary conditions must be supplied to prevent rotation about the z-axis and to prevent translation in the y-direction. The exact solution of the homogeneous (f = 0) general form of (10.2-1) is given in terms of hyperbolic sines and cosines. Based on Tong s Theorem, exact solutions at the nodes are obtained if such functions are used in a finite element model. Advanced elements of that type have been applied with excellent results. Only when there is no foundation support (k = 0) and no axial force (N = 0) will the simplified homogeneous solution of d 2 [EI d2 v dx 2 dx2] f(x) = 0 (10.2-3) be a cubic polynomial. All of the Hermite polynomials include at least the cubic. Therefore, a finite element model of (10.2-3) using a Hermite interpolation will always give exact values of v(x) at the nodes in the mesh, even if it is only approximate at other points in the element. When the model differential equation represents a beam then the following quantities are considered in engineering studies: EI v (x) = f(x) load per unit length, Differential equation EI v (x) = V(x) transverse shear force, Nonessential boundary condition EI v (x) = M(x) bending moment, Nonessential boundary condition v (x) = θ(x) slope, Essential boundary condition v(x) deflection, Essential boundary condition. There are six exact solution cases for the homogeneous portion of (10.2-1) that depend on the relative magnitudes of the coefficients EI, N, and k. Those solutions depend of the eigenvalue solution of the homogeneous equation and will not be considered in general in this section (eigen-problems are covered in Chapter 12). Of course, a numerical solution of that equation is also dependent on the relative values of those coefficients, but automatically finds the best approximate solution Integral form: To apply the Galerkin weak form the governing ODE is multiplied by v(x) and the integral over the length of the beam is set to zero. The highest (4-th) derivative term is always present, and needs to be integrated twice by parts to reduce the inter-element continuity requirement. That integral becomes L 0 I E = v d2 dx 2 (E(x)I(x) d2 v dx2) dx (10.3-1) 5

6 I E = [v d (E(x)I(x) d2 L v dx dx 2)] dv [ (E(x)I(x) d2 L v 0 dx dx 2)] 0 L 0 dx 2 + d 2 v (E(x)I(x) d2 v dx2) dx (10.3-2) The integration by parts brings the two possible nonessential boundary conditions, or reactions, into the integral form. Those terms are the transverse shear force, d(ei d 2 v dx 2 ) dx, and the moment, EI d 2 v dx 2. The first NBC term is the product of the displacement and the transverse shear force at the end points, while the second term is the product of the slope (angle) and the point moment at a the end points. Both of those products are definitions of mechanical work. Thus, the contributions from the fourth derivative term in the ODEs of (10.2-1) or (10.2-2), for N = 0, are I E = [v V(x)] L 0 [ dv M(x)] L + L d2v (E(x)I(x) d2 v dx 0 0 dx 2 dx2) dx. The second integral term from the axial force originally is non-symmetric because it also has unbalanced derivative orders. Creating a symmetric form requires integration by parts: L I N = v 0 ( N(x) d2 v dv dx2) dx = [v ( N(x) )] dx 0 L + dv L 0 dv N dx dx dx (10.3-3) For small deflections, the slope is approximately equal to its tangent. Thus, the boundary term N dv dx N y is the transverse (y-) component of the axial force in the deformed beam. The final integral form contains second derivatives (d 2 v dx 2 ) as its highest derivative term. The presence of second derivatives in the integral form means that calculus requires the elements to have inter-element continuity of the deflection and the slope, v(x) and θ(x) = v (x). Such elements are said to have an inter-element continuity of C 1. Therefore, each shared beam node must have two degrees of freedom (dof), at least. The Galerkin models for the other terms in (10.2-1) were covered in Chapter 8 and are now appended to give the complete governing Galerkin integral form of the beam-column equilibrium equation: I = [v { d (E(x)I(x) d2 v dv dx dx2) N(x) }] L dv dx [ (E(x)I(x) d2 L v 0 dx dx 2)] 0 L 0 dx 2 + d2 v (E(x)I(x) d2 v dx 2) dx + dv L dv N 0 dx dx L 0 dx dn dv dx dx dx L + vkv dx 0 L v 0 v dx L 0 vf dx = 0 (10.3-4) The first nonessential boundary condition states that when N 0 the consistent definition of the transverse shear force is V(x) = {d(ei(x) d 2 v dx 2 )/dx N(x) dv dx} V E V N. (10.3-5) Equation (10.3-4) must be satisfied along with any essential boundary conditions. The finite element model now has at least two degrees of freedom per node (n g = 2). At node number k (local or system) they will be denoted as v k and θ k. Likewise the element degrees 6

7 of freedom are locally numbered sequentially over the n n number of nodes per element to create a total of n i = n g n n independent degrees of freedom per element: δ et = [v 1 θ 1 v 2 θ 2 θ nn ] (10.3-6) Likewise, the point sources at any node are a transverse shear force, V k, and/or a point moment (couple), M k. The possible point load vector for an element is c P et = [V 1 M 1 V 2 M 2 M nn ] (10.3-7) It would typically be a mixture of reaction terms from the EBC and externally applied point sources (most of which are identically zero. The element generalized displacements are a subset of the n m system unknowns, δ e e δ: δ T = [v 1 θ 1 v 2 θ 2 v 3 θ 3 θ nm ] which correspond to a total of n d = n g n m independent equations in the system (before EBC) Element Arrays: As before, substituting the Hermite interpolation function for the beam s transverse deflection and slope: v(x) = H e (x) δ e and v (x) = dv dx = θ(x) = dh e dx δ e leads to the definitions of the element matrices. Expressed in matrix notation the equilibrium equation in (10.3-3) becomes: I = δ T c NBC + δ T S E δ + δ T S N δ + δ T S k δ δ T S w δ δ T c δ T c f = 0. This leads to the matrix system: [S E + S N + S w + S k ]δ = c NBC + c P + c + c f (10.4-1) which is assembled from the corresponding element matrices of: S E e = d2 T H(x) dx E e L e 2 I e d2 H(x) dx 2 dx, S E e B e (x) T L e EI e B e (x)dx, T N L e B e (x) d2 H(x), S dx 2 N e = dh(x) dx S e w = H(x) T dn e L dx e dh(x) dx, dx dh(x) dx, S dx k e = H(x) T k e H(x) dx, L e c e f = H(x) T f e (x) dx, c e L e = H(x) T e v L e dx (10.4-2) The new matrix, S N e, which is dependent on the axial force stress, and thus the axial stress, is usually called a geometric stiffness matrix in the finite element literature. The bending stiffness matrix is S E e, the foundation stiffness matrix is S k e, while the axial force gradient matrix has no common name. Referring to (10.3-5), the first row of the reaction/natural boundary condition vector, c NBC, contains the transverse shear {V E (0) V N (0)}, the second row contains the point bending 7

8 moment { M(0)}, the next-to-last row contains the transverse shear force { V E (L) + V N (L)}, and the last row contains the end point bending moment {M(L)}. Any distributed transverse load per unit length is usually defined by the value of that load input at the element nodes and approximated with a Lagrangian interpolation function. Assume that f e (x) = h w (x) f e. Then the resultant element load vector changes into an alternate rectangular integral: c e f = H(x) T h w (x) f e dx = [ H(x) T h w (x) dx] f e R e f e L e L e R e H(x) T h w (x) dx (10.4-3) L e n i n n = (n i 1)(1 n n ) Temperature changes along the length of the beam do not affect the transverse deflections or generalized forces. However, a temperature change, T, through the thickness, t e, from top to bottom does affect them. It is possible to include those thermal effects in a beam model. Thermal bending is not common, but the result of including it gives another load vector acting on the element: c α e = B e T L e EI e (x) α e T(x) e t(x) dx = αe T e EI e t e B e T L e (x) dx (10.4-4) where α e is the coefficient of thermal expansion of the material. The foundation stiffness matrix, S k e, has the form previously called a generalized mass matrix. If this model is later extended to be time dependent, then the literal mass matrix will be required. In this case it is m e = H(x) T ρ e A e H(x) dx (10.4-5) L e where ρ e is the mass density of the material, and A e is the cross-sectional area of the beam. Of course, the volume of the beam is A e L e and its total scalar mass is m e = ρ e A e L e C 1 Element models: The Summary of Chapter 2 gives the cubic (two-noded) and quintic (three-noded) Hermite C 1 line interpolations. The C 2 quintic (two-noded) interpolations are also given there. The cubic beam model is by far the most widely used beam element. However, being a cubic polynomial its third derivative (shear force) is constant along the length of the element. That requires a large number of elements to generate a reasonable estimate of the transverse shear force which is a key parameter in actual beam design. Therefore, it is recommended that a quintic (5-th degree) polynomial model be used since its quadratic third derivative will give the exact shear force in the vast majority of applications of beams. The element matrices for both of those C 1 Hermite interpolations are summarized here. If single two-node cubic C 1 beam element is in equilibrium then its matrix system is EI e [ L L 12 6L v 1 6L 4L 2 6L 2L 2 θ ] { L 12 6L v } + 2 6L 2L 2 6L 4L 2 θ 2 N e [ 30 L 36 3L 36 3L v 1 3L 4L 2 3L L 2 θ ] { L 36 3L v } 2 3L L 2 3L 4L 2 θ 2 8

9 dn dx 1 [ 60 = { 30 6L 30 6L 6L L 6L 30 L 2 ] { 6L 6L L 2 6L 0 V 1 M 1 } + V 2 M 2 L 60 [ v 1 θ 1 v } + 2 θ L 2L ] { f 1 } + v 9 21 f 2 2L 3L e L { 12 k e L 420 [ L 54 13L v 1 22L 4L 2 13L 3L 2 θ ] { L L v } 2 13L 3L 2 22L 4L 2 θ 2 6 L } + αe T e EI e 6 L t e { 0 1 } (10.5-1) 0 1 The rectangular load transfer matrix in the above equation is (10.4-2) with a scalar interpolation for a linear variation of the load per unit length between the two nodes. The four square matrices would be combined into a single square symmetric stiffness matrix. Note that an axial compression force (N e < 0) reduces the stiffness, and the extra support from a foundation increases the overall stiffness. Also, note that a temperature change through the thickness of the beam only causes only bending effects, not the transverse shear. If the recommended three-node quintic C 1 beam element is in equilibrium then its matrix system is dn dx EI 35L 3 + N 630 L + 1 1,260 k L 13,860 5,092 1,138L 3,584 1,138L 332L 2 896L 3, L 7,168 1,920L 320L 2 0 1, L 3,584 [ 242L 38L 2 896L 1,920L 1, L 320L 2 242L 38L 2 0 3, L 1,280L 2 1,920L 320L 2 1,920L 5,092 1,138L 320L 2 1,138L 332L 2 ] { 1, L 1, L L v 1 39 L 28 L 2 48 L 8 L 2 9L 5 L 2 θ 1 1, L 3, , L v L 8 L L L 8 L 2 θ L 1, L 1, L v 3 [ 9 L 5 L 2 48 L 8 L 2 39 L 28 L 2 ] { θ 3 } L L L v 1 46 L 0 32 L 8 L 2 14 L L 2 θ L L L v L 8 L L L 8 L 2 θ L L L v 3 [ 14 L L 2 32 L 8 L 2 46 L 0 ] { θ 3 } v 1 θ 1 v 2 θ 2 v 3 θ 3 } 2, L L L v L 8 L 2 88 L 12 L 2 29 L 3 L 2 θ L 5, L v L 12 L L L 12 L 2 θ L L 2, L v 3 [ 29 L 3 L 2 88 L 12 L L 8 L 2 ] { θ 3 } 9

10 = V 1 M 1 V 2 M 2 V 3 { M 3 } + L 420 [ L 4L L 0 8L L 3L] f 1 { f 2 f 3 } + v e L L { 7L} + αe T e EI e t e { 1} (10.5-2) The rectangular load transfer matrix in (10.4-2) is for a quadratic variation of the load per unit length between the three nodes. The Matlab symbolic integration of (10.4-3) to construct that rectangular matrix is given in Fig A constant transverse load per unit length (f) reduces the resultant load and moment vector to c f T = fl[98 7L L]/420, while a triangular load (f 1 = 0, f 2 = f 2, f 3 = f) reduces to c f T = fl[19 2L 112 8L 79 5L]/420. This and other line load conversions are symbolically calculated in Fig The associated mass matrix for both beam elements is like their elastic foundation matrix, except for the leading constant, k L, is replaced with the total mass of the beam element, m = ρ e A e L e. The associated mass matrix for the cubic element is L e m e = ρe A e L e 22L e L e 2 13L e 13L e 3L e L e 13L e 156 3L e 2 22L e ] (10.5-3) 22L e 4L e 2 The first three stiffness matrix calculations can be executed in scripts matrix_s_e_l3_c1.m, matrix_s_n_l3_c1.m and matrix_s_k_l3_c1.m. The rectangular transform matrix is included in matrix_c_f_l3_c1.m to accept three line-load values and build their resultant vector. The basic bending stiffness matrix is always required in any structure; the distributed line-load resultant is commonly needed. The elastic foundation stiffness is sometimes needed; as is the thermal bending matrix. The axial load geometric stiffness matrix is rarely needed, except for pipelines and buckling studies. The node point load vector is frequently needed for optional reaction recovery. Point sources are usually directly scattered to the system vector, but can be input at an element level. A common special case in well drilling and piping has the beam (pipe or drill string) at or near vertical. If β is the angle between the beam x-axis and the downward vertical direction (and gravity vector) and the beam weight per unit length is w then, the rate of change of the axial force is dn dx = w cos β and the two axial forces are related by N 2 = N 1 ± w L cos β. Then the combined effect of the axial load is for the quintic beam is 10

11 S N e = N 1 1,260 L 2, L 2, L L 32 L 50 L 2 64 L 24 L 2 32 L 5 L 2 2, L 3, L L 368 L 24 L L 256 L L 8 L 2 [ 14 L 5 L 32 L 8 L 2 46 L 6 L ] L L L + N 2 1,260 L L L L 46L 6 L 2 32 L 8 L 2 14 L 5 L L 3, L 2, L 112 L 8 L L 256 L L 24 L L 2, L 2, L [ 32 L 5 L 2 64 L 24 L 2 32 L 50 L 2 ] (10.5-4) 10.6 Classic beams: The analysis and design of beams has been important for at least two hundred years. In a current textbook on the Mechanics of Materials 97 of its 750 pages (13%) were devoted to the deflections and shear and moments in a beam. The above matrix system in (10.5-3) alone, or in combination with the derivatives of the C 1 interpolation functions, will give the same exact solutions for point forces, point couples, and line-loads that are constant, linear, or quadratic along the element. Such textbooks generally do not cover the influence of elastic foundations. If an elastic foundation is present the results of (10.4-3) are only approximate (because the analytic solution includes hyperbolic and trigonometric functions). Then, a finer mesh of three-noded elements is required. If that mesh is not fine enough then the shear force graph will show small jumps in the shear force at element interfaces within a single span. 11

12 Figure Symbolic integration to form the rectangular line-load conversion matrix Figure Symbolically computing the resultant source vector from line-loads. 12

13 Figure Symbolic solution of a quintic fixed-fixed beam with a triangular line load Now that each node has two degrees of freedom, the packed integer code that flags the essential boundary condition at each node can have the following packed values: 00 free joint, 10 v is given (a roller joint), 01 the slope given, or 11 both v and the slope are given (a fixed joint). It is misleading to just observe from Tong s theorem that for common line load distributions two cubic elements and a single quintic element can both give the exact deflections at the nodes and the exact reactions when using the same number of unknowns. The important items in designing beams are the moment distribution and the transverse shear force distribution along the beam. The moment and shear estimates from the classic cubic beam are linear and constant, respectively; while for the quintic element they are cubic and quadratic quantities. The shear force in a beam is very rarely constant, but it frequently varies quadratically along the beam. To obtain reliable shear design data with the cubic beam requires a very large number of elements along the beam. Better shear estimates can be obtained with a few quintic elements. In both cases, there still must be an element interface where any discontinuity in the data occurs. Figure shows a fixed-fixed beam span with a triangular line load and Fig graphs element 13

14 moment and shear force estimates. The classic cubic beam results are the dashed lines. The solid lines are the quintic (and exact) values of the moment and shear distributions. Figure Fixed-Fixed beam with a triangular line load (use Craig) Figure Non-dimensional moment (left) and shear for L2 (dashed) and L3 beams Figure again shows that the partitioning of the system matrices through the use of vector subscripts is quite useful in symbolic solutions. It is less useful in large numerical solutions because the vector subscripts create temporary hidden arrays (if memory permits) which are automatically deleted after their use. Figure shows detailed manipulations when all of the essential boundary conditions are zero. However, it is common for some of the EBC values to be given to be zero while others have non-zero values specified. For example, Fig shows a fixed-end beam where a support movement has raised the right end of the beam (or a settlement lowers the left end). The end displacement is an essential boundary condition. Whether other loadings are present or not, the known displacement of a support introduces other loads on the structure. To calculate what the reactions are due to a settlement (a non-zero EBC) the system matrices can be partitioned (via three vector subscripts) into regions of zero displacements, non-zero displacements and unknown displacements, where {r} denotes the reactions at EBCs: 14

15 S ff S fn S fz u free S nf S nn S nz { u not } = { u zero [ S zf S zn S zz] The top partition gives the free displacements the reactions at the non-zero EBCs are c free c not 0 r not u free = S ff 1 {c free + 0 S fn u not S fz 0} } + { } (10.6-1) c zero r zero r not = S nf u free + S nn u not + S nz 0 c not and the bottom partition yields the reactions at the zero EBCs: r zero = S zf u free + S zn u not + S zz 0 c zero. Figure Fixed-end beam with support motion In general, the non-zero EBC values in the partitioned arrays of (10.6-1) can be assigned to multiple degrees of freedom so the Not_0 vector subscript will also have several degree of freedom numbers, instead of one as shown in Fig For example, in that figure if the right support had also rotated counter-clockwise by an angle of θ then the vector subscripts become Is_0 = [1, 2]; Not_0 = [5, 6]; and Free = [3, 4]. Now, a new vector is required to specified the non-zero boundary condition value for each degree of freedom in the subscript array Not_0, say Not = [d θ]. Figure shows the full Matlab details for symbolically partitioning a matrix system into three groups of degrees of freedom. Here, the total number of degrees of freedom is only six, but for numerical calculations there is no theoretical limit on the size of the three vector subscript arrays that govern the solution of the system and the reaction recovery. 15

16 Figure Partitioning the displacements into three sets Example Given: For the beam in Fig , modify the Matlab script in Fig to symbolically compute the mid-span deflection and slope, the induced reactions and check the reaction values with statics. Solution: To easily find the mid-span deflection use a three-node quintic beam element (or two two-noded beam elements, since both methods give exact results). The system degrees of freedom are 1, 2; 3, 4; and 5, 6. Clearly, the center displacements (3, 4) are free, while the right vertical displacement (5) has a non-zero value. The other support degrees 16

17 of freedom (1, 2, 6) are zero. Those distinctions can be defined by vector subscripts, say Is_0 = [1, 2, 6]; Not_0 = [5]; and Free = [3, 4] that partition the matrix equations: S 3:4,3:4 S 3:4,5 S 3:4,(1,2,6) u 3:4 c 3:4 0 S 5,3:4 S 5,5 S 5(1,2,6), { u 5 } = { c 5 } + { r 5 } u (1,2,6) c (1,2,6) r (1,2,6) [ S (1,2,6),3:4, S (1,2,6),5 S (1,2,6),(1,2,6)] Solving for the mid-span displacements u 3:4 = S 3:4,3:4 1 {c 3:4 + 0 S 3:4,5 u 5 + 0}: { v 2 EI θ } = [ 2 35L 3 [7, ,280L 2]] { fl 14,700 {3, L } EI 35L 3 [ 3,584 d ] {d} = { 2 1,920L 3d 2L }} As outlined above, the reaction load at the non-zero EBC is found to be {r 5 } = {12 EI d L 3 }, which agrees with the upward right reaction force of Fig Likewise, the bottom partition gives the other exact reactions in that figure. Example Given: A fixed-fixed three-node beam with a triangular line-load, as shown in Fig The essential boundary conditions are v 1 = 0, θ 1 = 0, v 3 = 0, and θ 3 = 0. The external point force and moment at center node 2 are zero (V 2 = 0, M 2 = 0). Here, there is no axial force, N = 0, and no elastic foundation, k = 0 and v = 0, and no thermal load, α = 0. Determine the deflections and reactions using a single three-node (5-th degree) beam element. Solution: For a single element the equilibrium equations are given in (10.5-2). The middle two rows define the remaining unknown center point generalized displacements: After enforcing the EBCs: EI [7, L 3 0 1,280L 2] {v 2 θ } = { } + fl {3,920 14, L } + {0 0 }. Multiplying by the inverse of the square matrix gives the middle node solutions: { v 2 θ 2 } = 35L3 EI 7,168 [1 0 fl 0 1 1,280L 2 ] 14,700 {3, L } = 17 fl3 3,840 EI {5L 2 } which are the exact mid-span deflection and slope. Here, the reactions on the left are found from the first two rows of the equilibrium equations (since all the displacements are now known): EI 35 L 1,920L [ 3, L 320L 2 ] fl3 {5L 3,840 EI 2 } = { V 1 } + fl M { ,700 70L }, giving { V 1 } = fl M { L } Likewise, at the right end, utilizing the last two rows of the equilibrium equations gives the reactions at node 3: { V 3 M 3 } = fl 20 { 7 L }.

18 Both of the reactions are exact. Note that the net resultant force is fl/2, which is equal and opposite to the applied triangular shaped transverse load. The net external moment is zero. Static equilibrium is usually taught in undergraduate classes with Newton s Laws. You can use them to verify that the computed reactions do indeed satisfy that the sum of the forces is zero, and that the sum of the moments, taken at any reference point, is zero. The interpolated deflection of a quintic beam with these EBCs is v(x) = v 2 (16r 2 32r r 4 ) + θ 2 ( 8r r 3 40r r 5 )L Substituting the above values for the mid-point deflection and slope, the deflection estimate reduces to a fifth order polynomial v(x) = f L 4 r 2 (r 1) 2 (r + 2)/120EI = f L 4 [r 5 3r 3 + 2r 2 ]/120EI where r = x/l. Handbook solutions show that the exact displacement is the above fifth degree polynomial and therefore this single element model gives the exact deflection, slope, moment and transverse shear force everywhere along the length of the beam. The finite element solution does not utilize Newton s Laws, but the can be used to validate the results. This solution is executed symbolically, and the results checked, in Fig Example Given: Repeat Ex using two two-node cubic beam elements. Solution: The element lengths are both L e = L 2 and the line load values at the left and right elements are [0 f/2] and [f/2 f], respectively. From (10.6-1) the left element nodal resultant values are T c f = f L [ 9 2L/2 21 3L/2] / 240, and the right ones are T c f = f L [ 39 7L/2 51 8L/2] / 240, respectively. Adding the force rows (1 and 3) gives f L( )/240 = f L/2 which, as expected, is the total load caused by the variable line load. The two elements share degrees of freedom three and four so the assembled system is 12 6L/2 6L/2 4L 2 /4 EI 12 6L/2 ( L 3 2 ) 6L/2 2L 2 /4 0 0 [ L/2 6L/2 2L 2 /4 ( ) ( 6L/2 + 6L/2) ( 6L/2 + 6L/2) (4L 2 /4 + 4L 2 /4) 12 6L/2 6L/2 2L 2 /4 0 0 v θ L/2 v 2 6L/2 2L 2 /4 θ = L/2 v 3 6L/2 4L 2 /4 ] { θ 3 } V 1 M 1 V 2 M 2 V 3 { M 3 } + fl 240 { 9 2L/ L/2 + 7L/2 51 8L/2 } 18

19 Since the two ends are fixed (have zero deflections and slopes) the middle two rows define the remaining unknown center point generalized displacements: 8EI L 3 [ L 2] {v 2 θ 2 } = { 0 0 } + fl 240 {60 2L } + {0 0 } { v 2 θ 2 } = L3 8EI 24 [ L 2 ] fl 240 {60 2L } = fl3 3,840 EI {5L 2 } which again is exact, as expected. However, it is only exact at the nodes and is inexact along the length of each of the two elements. This model also gives the four exact end reactions as before Structural symmetry: Often a problem has a symmetry or anti-symmetry condition that allows half of the one-dimensional domain to be modeled. When an analyst uses that condition the edge of the omitted domain must be replaced by a new boundary condition. Figure showed the concepts of one-dimensional symmetry and anti-symmetry for a scalar unknown. A symmetry (mirror) plane is one where the domain and its material (coefficients in the ODE) are mirror images, as are the EBCs. For the symmetry condition half-of the line is meshed and at the point falling on the plane of symmetry the boundary conditions are enforced. When displacement vectors and (infinitesimal) rotations vectors are present the distinctions between symmetric and anti-symmetric problems is sketched in Fig Figure Symmetric (left) and anti-symmetric Hermite models For an anti-symmetry plane the domain and its material are mirror images, but the EBCs have equal and opposite changes in value, relative to the value at the point in the plane. The line is cut in half, and the removed material is replaced with an essential boundary condition at the point in the plane that the displacement components in the plane are zero, and the rotational component normal to the plane is zero. Thus, an anti-symmetry solution over half of the line gives the change in the solution value v(x), with respect to the plane value, v p, so the actual solution is v(x) = v p + v(x). On the other (omitted) half of the line the change is the solution has its sign reversed and the actuals solution is v(x) = v p v(x). (As the figure indicates, v p is usually zero for vector degrees of freedom.) When the computer memory is insufficient to solve a problem with symmetric geometry and a general load state it is possible to solve two half-size models by decomposing the general load into symmetric loads and anti-symmetric loads. This analysis approach is sketched in Fig Of course the two solutions must be combined in the post-processing to describe the deflections and stresses throughout the entire structure. Some commercial finite element systems 19

20 have the ability to automatically carry out the post-processing and to display the results as if the complete structure had been analyzed. Figure Combining a symmetry and an anti-symmetry model for a beam solution For larger structures there are more conditions associated with symmetry and anti-symmetry planes that can be used to reduce the problem size at the expense of solving two or more smaller problems, as shown in Fig For one-dimensional problems it is nice to cut the number of unknowns in half. For two- and three-dimensional problems it is often a necessity. Using either condition reduces the memory requirements by two, and the number of solution operations (and thus solution time) by a factor of about eight. Your computer will at some point become too small to solve real problems without using one or both of these conditions. 20

21 Figure Symmetry and anti-symmetry in vibration, buckling, and stress analysis Symmetric beam example 10.8 The Rayleigh quotient: In 1870 Lord Rayleigh showed that the natural frequency, ω, of vibration of any elastic system was related to the quotient of the strain-energy of the system 21

22 over the kinetic-energy of the system. Let u be the displaced shape of the body in a vibration mode. The strain-energy and kinetic-energy of an elastic system are U = 1 2 ut Ku KE = 1 2 u T Mu = ω2 2 ut Mu (10.8-1) because for simple harmonic motion the velocity is u = ω u. Therefore, the Rayleigh quotient gives the frequency of vibration for that mode as ω 2 = ut Ku u T Mu (radians/sec). (10.8-2) Note that the amplitude of the vibrating displacement shape, u, cancels out. Normally both the frequency, ω j, and the mode displacement shape, u j, are obtained together from an eigen-analysis (covered in Chapter 12) for the j-th mode of the system (1 j < ). That is a more complicated calculation than the static deflection calculation. For a given set of essential boundary conditions the static deflected shape should be a reasonable approximation of the shape of the first vibrational mode (j = 1). Lord Rayleigh showed that using a static deflected shape as an approximation of the first mode shape gives an upper bound estimate of the first (lowest) natural frequency of the system. The lowest vibrational frequency is often the most important one. Therefore, it is useful to substitute the displacement answers from a static structure into the Rayleigh quotient to estimate the structure s natural frequency. Ideally, the static load should be a reasonable approximation of how the mass is distributed through the structure. For a beam that means that the best natural frequency estimate, for a set of EBCs, would come from a static study with a uniform line load. Figure illustrates that in static solutions (to be used in the Rayleigh quotient), for general vibration modes (and for the lowest buckling mode) limited computer resources can force an analyst to carry out multiple symmetric and anti-symmetric boundary conditions to obtain the full range of solutions. The lowest eigenvalue (see Chapter 12) obtained from the four models in that figure will be the one that governs. Splitting a model in that fashion is not common for beam solutions, but it is not rare for two- or three-dimensional solids, so it is a good practice to learn how to use them in simple geometries. Example Given: The fixed-fixed beam of Example has a triangular line load and is approximated by the assembly of two cubic beam elements. Use its static deflection to estimate the natural frequency of the beam. Solution: Since most of the displacement components correspond to zero essential boundary conditions only the free degrees of freedom need to be used in the Rayleigh quotient. They are the partitions K(Free, Free), M(Free, Free), and u(free). The assembly of the free partition of the mass matrix is needed. From (10.6-1) with L e = L/2: M Free = ρe A e L e 420 [ ( ) ( 22Le + 22L e ) ( 22L e + 22L e ) (4L e2 + 4L e2 ) ] = ρal 420 [ L 2] The static deflections of u(free) = fl3 3,840 EI {5L 2 } 22

23 give the strain-energy and kinetic-energy as U = 4864 EI L and KE = 976 ρa L and the upper bound for the first natural frequency (in radians per second) is ω 1 = L 2 EI ρa < ω exact = L 2 EI ρa which is a 2.2% overestimate of the exact frequency. Using a uniform load to better approximate the mass distribution gives and a 1.6% overestimate. Example Given: Use the quintic fixed-fixed beam of Example with a uniform load to estimate the natural frequency of the beam. Solution: the partitions (see ) are K Free = EI [7, L 3 0 1,280L 2] M Free = ρal [5, , L 2] u Free = fl3 {L 384 EI 0 } and the upper bound for the first natural frequency (in radians per second) is ω 1 = L 2 EI which is a 0.3% overestimate of the exact frequency. ρa > ω exact = L 2 EI ρa 10.9 Multiple span beams: Many beams have more than a single span. Figure shows a typical two-span beam. The first span has a constant uniform line-load, while the second span has point load at the two-thirds point. It is classified as a statically indeterminate beam since there are six unknown reactions at the supports. Points A and B rest on rollers that prevent any vertical displacement and fixed-end C prevents any vertical displacement or rotation. The EBCs at those three points introduce discontinuities in the shear forces and the end moment. Therefore, element interfaces must occur there. The point load also causes a shear discontinuity so it too must be at an element interface. To get exact shear and moments under the uniform line load requires the three-noded quintic beam. Using that same beam element on both sides of the point load leads to a mesh with seven nodes and three elements. Figure Two span, constant EI beam 23

24 Note that the beam has a constant EI product value. Here E is the material modulus of elasticity of the material and I is the geometric moment of inertia of the cross-section of the beam. Often, the inertia is not known to begin with since it must be selected based on the values of the extreme moment and/or shear values in the system. It is a common practice to set the EI product to unity (with the proper units) and solve the problem with that value since the results are directly proportional to that product. Figure Deflections for the multi-span beam of Fig

25 Figure Deflection (top), moment, and shear diagrams for Fig To begin illustrating how to automate a multi-span beam analysis the Matlab script in Fig was developed to determine the deflections and rotations for the indeterminate system in Fig It does not have the usual for loop over all elements, but addresses each element sequentially (which is a much less efficient process.) The stiffness matrices and line-load resultant arrays are in (10.3-1). However, each of the beams has a different length and a different connectivity list. The vertical point load will be inserted directly into the system load vector, c, in the row corresponding to the vertical displacement at that node. The system equation number for degree of freedom j at node number N is row (j) = n g (N 1) + j, 1 j n g. For this beam n g = 2 with j = 1 corresponding to a vertical displacement and j = 2 being the small rotation angle (in radians) at the node. For the point load the equation number is 2*(5-1) +1 = 9. The required script is too long to fit a single page so its major stages are presented in Fig a and 3b. Figure a has comments to summarize the problem, manually sets several of the controlling integers, and manually sets the spatial coordinates and the mesh connectivity list. The last portion allocates the memory required for the major arrays to insure an efficient execution. Beginning with defining the vector subscripts for the unknown degrees of freedom, and the zero ones, Fig b assembles the point load, and manually loops over each element creates their arrays and scatters them. For each element in Fig b, the connecting nodes are gathered and used to get the nodal x-coordinates in order to compute the element length. The connections are used again with the control integers to generate the system degree of freedom numbers for the element. That is done with the get_element_index.m script given in Fig That script is always used for each element being assembled. All three elements use the script matrix_s_e_l3_c1.m to form their bending stiffness. Only the first element uses the script matrix_c_f_l3_c1.m to form its resultant due to the constant line-load. Each element array is scattered into the system matrices using its vector subscript, index. When the assembly process is complete, the partitioned system matrices are solved for the unknown displacements. Then all of the displacements are used to calculate the reactions at the essential boundary conditions. Finally, selected results are graphed. The short script in Fig showed that there are several actions that must be completed to fully automate a finite element analysis. Almost all studies require the input of the location and value of all essential boundary conditions. The library of scripts includes get_point_sources.m for that purpose. It reads the text file msh_load_pt.txt which has three columns: the node number 25

26 where the EBC occurs ( n m ), the degree of freedom direction ( n g ), and the value assigned to that generalized force. The user must always specify the coordinates of all of the nodes. Those node location inputs are read by script get_mesh_nodes.m which reads those data from the sequential text file msh_bc_xyz.txt. The number of lines in that file defines the total number of nodes in the mesh, n m. Every application must also specify the node connection lists of all of the element types (volumes, areas, lines, and/or points). They are read by script get_mesh_elements.m from the user supplied sequential text file msh_typ_nodes.txt. The number of lines in that file defines the total number of elements in the mesh, n e and the number of different types of elements, n t. The provided general finite element software system allows a mixture of compatible element types to be input. For example, it may contain a mixture of element types; say quadrilaterals and triangles and/or line elements. For a mixed mesh, the connection list must be padded with zeros to assure that the input text file has a constant number of columns (in Matlab, but not in Fortran). 26

27 Figure a Beam sketch, controls, basic data, and memory allocation 27

28 Figure b Assemble beam matrices, solve for displacements, and recover reactions Most problems also have specified properties, or coefficients in the ODE, which are application dependent. Their order of input depends on the person that writes the application script. The script get_mesh_properties.m reads the element or type properties from the text file msh_properties.txt and counts the number of properties n_vals (columns of data) and it counts the number of rows of data, n_mats. The number of rows of properties must equal either the number of elements or the number of different types of elements. A mixed mesh can also have different properties for each element type. When needed, the properties list for an element type must be padded with zeros to assure that the input text file has a constant number of columns. Figure shows typical numerical data that would be required had the two-span beam in Fig been solved numerically using the supplied general finite element library. Figure Numerical data equivalent to the beam in Fig for FEA library BOEF without axial load: The presence of an elastic foundation is not uncommon and significantly changes the solution to the homogeneous differential equation and its coupled beam structural response. Simple cases of beams on elastic foundations are known for infinite, 28

29 semi-infinite, and finite lengths. Typical responses of such beams will be illustrated through two numerical examples. In these two examples there is no reversal of the beam deflection direction. In other words, the foundation everywhere pushes back against the beam. It is not unusual for the beam to have a segment of its length that does reverse direction. Then part of the foundation is theoretically pulling on the beam. If the foundation is soil, then a gap would develop there instead of having the soil pull on the beam. That means the model would have to be changed by putting an element interface at the liftoff point, and the prior foundation stiffness removed for the new element. An accurate location of a lift-off-point can be a trial and error process. A cantilever BOEF with a triangular line load in Fig has an exact solution that consists of several products of hyperbolic regular trigonometric functions like cosh λx cos λx 4 where the constant λ = k 4EI is one of the common measures of the relative foundation to bending stiffness measures, and which has the units of one over length. The product λl generally classifies the beam as: short λl < π 4, medium π 4 λl π, or long λl > π. Of course, all of the exact expressions for the solution derivatives (slope, moment, and transverse shear force) involve similar complicated relations. Due to the complicated response of BOEF models a wise strategy is to refine the mesh until no clear gaps occur in the shear force (except of course at external point forces). Here, the classic cubic beam interpolation results for the deflections are compared to the quintic beam interpolations. The summary below shows that two three-noded beam members give better deflection results than four of the classic two-noded beam elements. Figure A cantilever BOEF with a triangular line load Table Cantilever BOEF with Triangular Load (f = 10 lb./in, E = 30e6 psi, I = 0.1 in 4, k = 1 lb./in) X Exact v L2_C1 1 Elem L2_C1 2 Elem L2_C1 4 Elem L3_C1 1 Elem L3_C1 2 Elem L3_C1 4 Elem However, the real comparison comes in examining the moments and shear, which are used to design beams. The graphs of those quantities are given in Fig For this type of loading, any discontinuity in the graph occurs at an element interface and indicates that the mesh is too coarse. The cubic four element mesh graph is on the left of that figure. Even the moment plot has such jumps indicated by the circled locations. The quintic two element mesh graph for the moment and shear agree with the complicated analytic solution to at least three significant figures. Note that at the free tip on the cantilever the quintic beam estimates go to zero, as 29

30 required by equilibrium, but the cubic elements do not. Analysts must use many cubic elements to hope to come close to the correct moment and shear needed for design purposes. It is fairly easy to get good deflections. Figure shows a more practical BOEF where the exact analytic solution has been given. Since there are discontinuous loadings any finite element model must split this beam into at least four spans. The first runs from the left end to the point load. The second covers the gap from the point load to the beginning of the line load. The third spans the length of the line load, and the last one covers from the end of the line load to the right end of the beam. The pressure distribution shown in that figure is actually computed from the final Figure Comparison of cubic (left) and quintic moment and shear in cantilever BOEF deflection as p = k v in a post-solution calculation and is proportional to the deformed shape of the beam. Also, note that there is no change in sign of the pressure or the corresponding deflections so no lift-off corrections need to be considered. The elastic modulus of this wooden BOEF is E = 1.5e6 psi, the moment of inertia is I = in. 4, and the foundation modulus is k = b k 0 = 10 in. 200 lb./in. 3 = 2e3lb./in. 2. The non-dimensional reference length for this beam 4 is λl = k 4EI L = 3.57 > π, which classifies it as a long beam. Generally, a point source at 30

31 one end of a long beam on an elastic foundation does not affect the deflection at the other end because the foundation dampens out its bending effect. This beam was run with both the cubic and quintic beam elements using the same number of nodes in the mesh. Each span had a mid-point node added. Thus, there are two cubic elements per span compared to one quintic element per span. Both models yield essentially the same displacements of the beam. The differences in the solutions again become clear in the moment and shear force evaluations. The moment and shear graphs for a nine node mesh are shown in Figs and -5, respectively. The cubic element mesh underestimated the maximum shear force by about 42%. Doubling the number of quintic elements only changed the shear results only in the fourth significant figure (too small to see). Thus, that quintic element mesh is probably sufficient. Doubling the number of cubic elements in a 17 node mesh gives the moment and shear results in Fig It is still possible to spot jumps in the moment graph. Figure BOEF with changing loadings Figure Moments from cubic (left) and quintic BOEF elements (10 nodes) 31